Let ℝ∗
be the group of nonzero real numbers with multiplication, and let
𝐻 = {𝑥 ∈ ℝ∗
| 𝑥
𝑛
∈ ℚ for some 𝑛 ∈ ℕ>0}.
1. Show that 𝐻 is a subgroup of ℝ∗
, and that it is normal.

Let's break down the problem of showing that \( H = \{ x \in \mathbb{R}^* \mid x^n \in \mathbb{Q} \text{ for some } n \in \mathbb{N}_{>0} \} \) is a subgroup of \(\mathbb{R}^*\) (the group of nonzero real numbers under multiplication) and that it is a normal subgroup.

### Step 1: Show that \( H \) is a subgroup of \(\mathbb{R}^*\).

To show that \( H \) is a subgroup, we need to verify the following three properties:
1. **Closure:** If \( x, y \in H \), then \( xy \in H \).
2. **Identity:** The identity element \( e \) of \(\mathbb{R}^*\) (which is 1) must be in \( H \).
3. **Inverses:** If \( x \in H \), then \( x^{-1} \in H \).

#### 1. Closure
Let \( x, y \in H \). By definition of \( H \), there exist positive integers \( n \) and \( m \) such that \( x^n \in \mathbb{Q} \) and \( y^m \in \mathbb{Q} \).

Consider the element \( xy \in \mathbb{R}^* \). We need to show that \( (xy)^k \in \mathbb{Q} \) for some \( k \in \mathbb{N}_{>0} \).

Observe:
\[
(xy)^{nm} = x^{nm}y^{nm} = (x^n)^m(y^m)^n
\]
Since \( x^n \in \mathbb{Q} \) and \( y^m \in \mathbb{Q} \), we have:
\[
(x^n)^m \in \mathbb{Q} \quad \text{and} \quad (y^m)^n \in \mathbb{Q}
\]
Thus, \( (xy)^{nm} \in \mathbb{Q} \). Therefore, \( xy \in H \), proving that \( H \) is closed under multiplication.

#### 2. Identity
The identity element in \( \mathbb{R}^* \) is \( 1 \). Since \( 1^n = 1 \in \mathbb{Q} \) for any \( n \in \mathbb{N}_{>0} \), we have \( 1 \in H \).

#### 3. Inverses
Let \( x \in H \). Then there exists some \( n \in \mathbb{N}_{>0} \) such that \( x^n \in \mathbb{Q} \).

Consider the inverse element \( x^{-1} \in \mathbb{R}^* \). We need to show that \( (x^{-1})^k \in \mathbb{Q} \) for some \( k \in \mathbb{N}_{>0} \).

Observe:
\[
(x^{-1})^n = (x^n)^{-1}
\]
Since \( x^n \in \mathbb{Q} \) and \( \mathbb{Q} \) is closed under inversion, \( (x^n)^{-1} \in \mathbb{Q} \). Therefore, \( x^{-1} \in H \).

Having verified closure, identity, and inverses, \( H \) is indeed a subgroup of \( \mathbb{R}^* \).

### Step 2: Show that \( H \) is a normal subgroup of \(\mathbb{R}^*\).

To prove that \( H \) is normal, we need to show that for any \( g \in \mathbb{R}^* \) and any \( h \in H \), the element \( g^{-1}hg \) is also in \( H \). In other words, \( gHg^{-1} \subseteq H \).

Let \( h \in H \) and \( g \in \mathbb{R}^* \). Since \( h \in H \), there exists some \( n \in \mathbb{N}_{>0} \) such that \( h^n \in \mathbb{Q} \).

Consider the conjugate \( g^{-1}hg \):
\[
(g^{-1}hg)^n = g^{-1}h^n g
\]
Since \( h^n \in \mathbb{Q} \), and \( g^{-1}\mathbb{Q}g = \mathbb{Q} \) (because \( \mathbb{Q} \) is a normal subgroup of \( \mathbb{R}^* \), the multiplication of rationals by any nonzero real number stays within the rationals), we have:
\[
g^{-1}h^n g \in \mathbb{Q}
\]
Therefore, \( (g^{-1}hg)^n \in \mathbb{Q} \), implying \( g^{-1}hg \in H \).

Since \( g^{-1}hg \in H \) for all \( g \in \mathbb{R}^* \) and \( h \in H \), the subgroup \( H \) is normal in \( \mathbb{R}^* \).

---

Would you like more details on any part of the proof, or do you have further questions? Here are 5 related questions to explore:

1. Can you describe an example of an element in \( H \) that is not in \( \mathbb{Q}^* \)?
2. What is the intersection of \( H \) with the subgroup \( \mathbb{Q}^* \subseteq \mathbb{R}^* \)?
3. What is the quotient group \( \mathbb{R}^* / H \), and how can it be described?
4. How does this proof generalize to other fields or rings?
5. How would you verify normality for a general subgroup of a group?

**Tip:** Remember that normal subgroups are precisely the subgroups that are invariant under conjugation by any element of the group, which plays a key role in constructing quotient groups.

Explore the proof that H, defined as the set of nonzero real numbers whose powers are rational, is a subgroup of the group of nonzero real numbers under multiplication. Understand why H meets the closure, identity, and inverse criteria for subgrouphood, and discover the proof of its normality within the group.

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